Question

15.0 g of ice cubes at 0.0°C are combined with 150. g of liquid water at 70.0°C in a coffee cup calorimeter. Calculate the final temperature reached, assuming no heat loss or gain from the surroundings. (The specific heat capacity of H2O(l), cs = 4.18 J/gC; H2O(s)  H2O(l) H = 6.02 kJ/mol)

Answer 1

Answer : The final temperature will be $66.7^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of ice = $2.09J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of ice = 15.0 g

$m_2$ = mass of water = 150 g

$T_f$ = final temperature = ?

$T_1$ = initial temperature of ice = $0.0^oC$

$T_2$ = initial temperature of water = $70.0^oC$

Now put all the given values in the above formula, we get

$15.0g\times 2.09J/g^oC\times (T_f-0.0)^oC=-150g\times 4.18J/g^oC\times (T_f-70.0)^oC$

$T_f=66.7^oC$

Therefore, the final temperature will be $66.7^oC$

Answer 2

This question involves the concepts of the law of conservation of energy, thermal equilibrium, and specific heat.

The final temperature reached is "58.33 °C".

According to the law of conservation of energy, in this situation:

Heat Released by Liquid Water = Heat Gained by Ice Cubes

Using, the formulae of specific heat to find out the final temperature at thermal equilibrium:

$m_iC_i\Delta T_i=m_wC_w\Delta T_w$

where,

$m_i$ = mass of ice = 15 g

$m_w$ = mass of liquid water = 150 g

$C_i$ = specific heat of solid ice = 2.093 J/g°C

$C_w$ = specific heat of liquid water = 4.18 J/g°C

$\Delta T_i$ = chage in temperature of ice = T - 0°C

$\Delta T_w$ = chage in temperature of water = 70°C - T

T = Final Temperature = ?

Therefore,

$(15\ g)(4.18\ J/g^oC)(T-0\ ^oC)=(150\ g)(2.09\ J/g^oC)(70^oC-T)\\62.7\ T + 313.5\ T=21945\\\\T = \frac{21945\ J}{376.2\ J/^oC}$

T = 58.33 °C

Learn more about thermal equilibrium here:

https://brainly.com/question/14556352?referrer=searchResults

The attached picture shows thermal equilibrium.